The Effects Copper Sulphate has on Catalase.

        Substrate

Hypothesis:

As you increase the concentration of copper sulphate, the gas being produce will decrease because the inhibitors have got a higher concentration then the substrate, which means that the substrate will not be able to bind with the active site because the inhibitor has got the same shape as the substrate and has connected to the active site ().

Prediction:

I predict that as you increase the concentration of copper sulphate, the rate of oxygen produced is decreased. This is because more inhibitors are occupying the active sites, which means that the enzyme cannot bind to the active site to do and complete its function ().

So if there is a low concentration of copper sulphate, then the amount of oxygen produced will increase because the substrate, which is hydrogen peroxide will have a higher concentration then the inhibitors, which is copper substrate, which means that the substrate will be able to bind to the active site and do and complete its function.

Method:

Apparatus –

I will be using the following apparatus:

• 3 plastic measuring cylinders

2 measuring cylinders will be used for measuring the solutions and the water and the other measuring cylinder will be used for to collect the gas.

I will be using measuring cylinders because this equipment is very easy to measure out solutions and to measure the amount of gas collected.

• 8 beakers

I will be using 8-glass beaker because the beakers were only the right and suitable equipment that I had to store the solutions that I had measured out.

• A clamp stand

I will be using a clamp stand because this is the only equipment that is big enough to hold my measuring cylinder in the bowl full of water.

• Plastic bowl

I will be using this because the bowl is big enough to collect the gas in the measuring cylinder.

• Conical flask

I will be using this because this is the only piece of equipment, which is available to me that will fit onto a delivery tube.

• Delivery tube

I will be using a delivery tube because this is the only equipment that is accurate enough to carry the collected gas from the conical flask to the measuring cylinder, without losing any.

• Digital stopwatch

  I will be using this because this is an accurate way of timing the investigation.

I will be using all of the apparatus above to do my investigation.

Setting up the apparatus –

I will be setting up the apparatus shown in the diagram.

• I am going to fill up the bowl with water until the bowl is nearly full.

        Nearly full

        Plastic bowl full of water

• I will then fill the measuring cylinder up with water to the top. I will then cover the top of the measuring cylinder with the palm of my hand, so no waterfalls out turn it upside down and then place it into the water. When the cylinder is in the water, I will take my hand away from it.

        

• Then use a clamp stand to hold the cylinder, which is placed in the bowl.

• Then I will put all the chemicals I am using into a conical flask and then connect a delivery tube from the conical flask to the measuring cylinder. I will then start timing straight away.

Concentration –

The volume of copper sulphate and water for the each end concentration are:

Method – step by step –

1. I will set up the apparatus shown in the diagram

2. I will then measure out all of the concentrations I need to do the experiment (see table 1 to find out how to make 0.25m, 0.5m, 0.75 and 1m)

3. I will then use the following amounts of solution to do one experiment:

I used a measuring cylinder to measure all of these solutions out.

4. I then poured all of the solutions into the conical flask, quickly connected the delivery tube up and then started timing for 5 minutes. I did it quickly so I do not lose a lot of gas.

5. After every minute I will record the amount of gas collected into my results table

6. After the 5 minutes are up, I will wash out the conical flask and start setting up for the next experiment

7. I will do each replicate 5 times to gain a reliable average.

Safety –

In this investigation, I will be wearing goggles and an apron. This is because we will be using some dangerous chemicals, which are very harmful, so if you got them in your eye, they could cause serious damage.

I will be wearing an apron because I would not like anything split on to my clothes, because it might not come off.

Variables -

In this investigation, the variables that affect the activity of the enzyme, catalase, were considered and controlled so that they would not disrupt the success of the experiment.

• Temperature:

As temperature increases, molecules move faster (kinetic theory). In an enzyme catalysed reaction, such as the decomposition of hydrogen peroxide, this increases the rate at which the enzyme and substrate molecules meet and therefore the rate at which the products are formed.

As the temperature continues to rise, however, the hydrogen and ionic bonds, which hold the enzyme molecules in shape, are broken. If the molecular structure is disrupted, the enzyme ceases to function as the active site no longer accommodates the substrate. The enzyme is denatured.

To control this variable, the temperature was maintained at a fairly constant level that allowed the enzyme to work effectively (room temperature, approximately 23ºC). This was achieved by using a test tube rack and tongs to handle the apparatus so that the heat from my hands did not affect the Catalase.

• pH:

Any change in pH affects the ionic and hydrogen bonding in an enzyme and so alters it shape. Each enzyme has an optimum pH at which its active site best fits the substrate. Variation either side of pH results in denaturation of the enzyme and a slower rate of reaction.

In this experiment, the pH was kept constant using a pH 7 buffer, selected to maintain a pH level suited to the enzyme by being equal to the natural environment of the enzyme (potato tissue).

• Substrate Concentration:

When there is an excess of enzyme molecules, an increase in the substrate concentration, produces a corresponding increase in the rate of reaction. If there are sufficient substrate molecules to occupy all of the enzymes´ active sites, the rate of reaction is unaffected by further increases in substrate concentration as the enzymes are unable to break down the greater quantity of substrate.

• Inhibition:

Inhibitors compete with the substrate for the active sites of the enzyme (competitive inhibitors) or attach them to the enzyme, altering the shape of the active site so that the substrate is unable to occupy it and the enzyme cannot function (non-competitive inhibitors). Inhibitors therefore slow the rate of reaction.

• Enzyme cofactors:

Cofactors are none protein substances which influence the functioning of enzymes. They include activators that are essential for the activation of some enzymes. Coenzymes also influence the functioning of enzymes although are not bonded to the enzyme.
Unless enzyme cofactors were present in the potato tissue containing the Catalase, they were not included in this investigation and therefore would not have affected the rate of reaction and the results of this experiment.

How I will be obtaining my results:

I will be gaining all my results to 2 decimal places. For the volume of the gas, I will record it to the nearest half or whole number.

I will also do each experiment 5 times in order to gain an average for each replicate. When the 5 minutes are up, I will be able to go over up to 10 seconds.

Changes from preliminary work:

The changes that I did from the preliminary work are as follows:

• Instead of using a syringe in my experiment to collect the gas, which was oxygen, I decided to use a plastic measuring cylinder tipped upside down in a bowl full of water. I decided to use this method because I was more familiar with it and it was easier for me to use. 
• Instead of using a potato as my main source of catalase, I decided to use neat catalase. This is because the potato solutions oxidised and turned blacked over night, which meant that I had to make more potato solution. This wasted a lot of time, so I decided to use neat catalase.
• Instead of doing each molar 3 times, I decided to do it 5 times to give me a better average of each molar.
• In my preliminary work, I used a water bath but then I decided not to use it because it denatured my catalase

Results tables:

The following results are the results from my experiment, in which I used neat catalase as my main source of catalase instead of the potato solution:

Conclusion:

In this investigation, I have found out that by adding copper sulphate to hydrogen peroxide and catalase, them amount of gas that is being produced reduces. This is because throughout this experiment, copper sulphate is acting as an inhibitor and is trying to bind to the active site before the hydrogen peroxide.

As you increase the molar, the copper sulphate gains a higher concentration than the hydrogen peroxide, which means that the copper sulphate is stronger than hydrogen peroxide, so it binds to the active site and reduces the amount of gas produced.

You can see this in my results table above. As the molar increases, the amount of gas being produced decreases.

0m to 0.25m – 

As you can see here, them amount of oxygen being produced reduced, but not that much. The amount of oxygen reduced between these two molars is 0.56cm3.

From here, we can see that the inhibitor, which is hydrogen peroxide has started to bind to active sites but not to a lot because the copper sulphate has still got more concentration then hydrogen peroxide.  

0.25m to 0.5m –

As you can see here, the amount of oxygen has still decreased but now it has decreased a lot from the first one. The amount of oxygen reduced between these two molars is 2.68cm3.

This shows that the hydrogen peroxide has started to gain more concentration then the substrate, which is copper sulphate and now has started to bind to active sites even more to reduce the amount of oxygen being produced.

0.5m to 0.75m –

As you can see here, the amount of oxygen produced has decreased a lot. The amount of oxygen reduced between these two molars is 37.84cm3.

This shows that now the hydrogen peroxide has more concentration then the copper sulphate, which means that the hydrogen peroxide will be binding to more active sites than the copper sulphate, reducing the function of the copper sulphate. 

0.75m to 1m –

As you can see here, the amount of oxygen has still reduced but not that much from the molars before. The amount of oxygen reduced between these two molars is 21.28cm3.

This shows that now the hydrogen peroxide has double the amount of concentration then copper sulphate, which means that now the hydrogen peroxide will be binding to even more active sites, reducing the function of the copper sulphate.

You can see the rest of my results and the pattern of my results in my results table above.

Evaluation-

From all of this investigation, I have found out that as you increase the molar (0m to 1m) the amount of oxygen produced reduces.

In this investigation, I did everything according to the method I wrote. I did add nothing while I was doing my experiment to the method nor take something out of the method.  I did change things from my preliminary, which I have also stated in my method.

The one main problem I came across to while I was doing my experiment was that went between the time when I put the solutions into the conical flask and I put the bung on, some gas produced would have escaped, which meant that I did not record all the gas produced. I tried to find a solution to overcome this. I researched this in several textbooks and Internet sites (see reference section at the end of this coursework) but I could not find anything.

The equipments that I stated in the method were all used. The equipments were perfect for this experiment especially for collecting the gas produced and measuring out solutions.

While I was looking through my results table and finding the averages, I found out that the 0.25m started off slower than the 0m but when it started to produce oxygen, it produced the highest level. This was a surprise to me because at this time, the hydrogen peroxide would have been binding to the active sites more than 0m but not as much as the copper sulphate, which meant that the gas being produced would have reduced and not exceed 0m but it did. But when I calculated the averages, the 0m average was higher than the 0.25m average and not that much.

If I had to do this experiment again, the things I would change or improve are as follows:

• I would do my experiment with the potato solution and then compare my results with the neat catalase. I know that the potato solution oxidises and turns black when left over night, so I would try to do the experiment in one day.
• In this experiment I was using 0m, 0.25m, 0.5m, 0.75m and 1m, if I had to do the experiment again, I would use different types of molars, e.g. 0m, 0.1m, 0.2m, 0.3m, 0.4m, 0.5m, 0.6m, 0.7m, 0.8m, 0.9m and 1m.

References

The following books were used to search up information about enzymes and copper sulphate, to help me with my investigation:

Biology – A functional Approach

2nd Edition

By M.B.V Roberts and T.J King

Biology 1

Cambridge Advanced Series

Endorsed by OCR

Series Editor – Mary Jones

Practice in Biology – Progressive Questions for AS and A level

Hodder & Stoughton Educational

By Sally Morgan

_______________________________________________________________________

The following websites were used to search up information about enzymes and copper sulphate, to help me with my investigation:

The following web sites are search engines, which allowed me to search up particular words.